09-02-2013
Date format in Bash Script
Hi Experts,
We get "Day" of a month in a variable, so how to make date of out it?
To make more sense
if my variable $DAY contains "12" and month and year will be current date (as of today)
so I want to see as output as 2013-09-12.
How can I achive this bash script??
Any help is highly appreciated.
Thank you
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LEARN ABOUT PHP
cal_to_jd
CAL_TO_JD(3) 1 CAL_TO_JD(3)
cal_to_jd - Converts from a supported calendar to Julian Day Count
SYNOPSIS
int cal_to_jd (int $calendar, int $month, int $day, int $year)
DESCRIPTION
cal_to_jd(3) calculates the Julian day count for a date in the specified $calendar. Supported $calendars are CAL_GREGORIAN, CAL_JULIAN,
CAL_JEWISH and CAL_FRENCH.
PARAMETERS
o $calendar
- Calendar to convert from, one of CAL_GREGORIAN, CAL_JULIAN, CAL_JEWISH or CAL_FRENCH.
o $month
- The month as a number, the valid range depends on the $calendar
o $day
- The day as a number, the valid range depends on the $calendar
o $year
- The year as a number, the valid range depends on the $calendar
RETURN VALUES
A Julian Day number.
SEE ALSO
cal_from_jd(3), frenchtojd(3), gregoriantojd(3), jewishtojd(3), juliantojd(3), unixtojd(3).
PHP Documentation Group CAL_TO_JD(3)